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kleraudio
10-11-2004, 03:33 AM
Hey everyone, had a quick question. i was having coffee w/ a friend tonight and we got to talking about odds and the math behind it. now im not a math whiz but we were trying to figure out the odds of getting any pocket pair. i got to 16:1 or 1 in 17 or 5.9%. not sure if thats right or not. say it is though, how do you figure out the odds that someone else at the table has a pocket pair as well (in the same hand of course). do you double it. would it be 32:1? I am just starting to get into this math part of poker and its alot of fun. your help is greatly appreciated.

garyc8
10-11-2004, 07:12 AM
1 in 17 is correct.
The odds of someone else having a pocket pair is a little more difficult. Each player at the table also has about a 1 in 17 chance of getting a pocket pair. (The two cards in your hand may change that fractionally; I'm not sure, but let's use 1 in 17.) That means each player has a 16 in 17 chance of NOT having a pair.
The chance that noone else has a pp is computed thus:
16*#of opponents/17*#of opponents.

This means that in a 10 handed game the chance that no other player holds a pp (whether you do or not) is 57.95%
So naturally, the chance that another player has a pp is 42.05%.

kleraudio
10-11-2004, 03:07 PM
hey cleveland guy, if i do 16*9(number of opp.)/ 17*9 i get 94%. where have i gone wrong??

Thanks for the help in advance

garyc8
10-11-2004, 07:36 PM
Sorry I misexpressed myself. It's 16 raise to the ninth power (nine opponents)/ 17 raise to the ninth power.
16*16*16*16*16*16*16*16*16/17*17*17*17*17*17*17*17*17

KingMarc
10-11-2004, 07:47 PM
As crappy as the book is, Ken Warren's Winning Texas Hold Em has a great section on probablity with everything, including pocket pairs.

For example, when you have KK, there is a 1 in 23 chance someone has AA.

BruceZ
10-11-2004, 10:00 PM
[ QUOTE ]
As crappy as the book is, Ken Warren's Winning Texas Hold Em has a great section on probablity with everything, including pocket pairs.

For example, when you have KK, there is a 1 in 23 chance someone has AA.

[/ QUOTE ]

I agree that Warren's probability tables are as comprehensive as any I have seen; however, I found an error with the table titled "Odds That Someone Holds a Better Hand than You Preflop". I have the January 1997 printing, and this table is on p. 193. The columns labeled "Number of players dealt in the hand" ranging from 8-10 should actually read "Number of opponents dealt in the hand". Alternatively, the text can remain the same, but then the numbering must change from 9-11 players.

For example, as you quoted, when you hold KK, he gives odds that someone holds AA as 21.8-1 for 9 players. This would be exactly correct for 9 opponents (10 players) using the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=417 383) 9*6/C(50,2) - C(9,2)/C(50,4) = 1 in 22.8 or 21.8-1. I have checked several others, and they were all exactly correct except for this error, so the table is still completely useable as long as this error is understood. If anyone has a newer edition, let me know if this has been corrected.

Of course, there is also a more obvious error with this table. It is completely bogus to list suits the way he does. For example, it's not the odds that someone holds Ah As, but any AA.

uuDevil
10-11-2004, 11:33 PM
Hi BruceZ,

Thanks for pointing out these errors. I have the 5th (Feb. 2002) printing and this table still has the problems you mention.

kleraudio
10-12-2004, 07:01 PM
is there any other way besides that equation to figure out that if you hold KK someone else holds AA. i know to get AA is is 220-1 (or any other specific pocket pair), why is it now 22-1 just because you hold KK?

Thanks